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Yes, same!

@@consaidercordo3770 thanks

You're not alone LOL

Yes In cylindrical PV, the hoop stress is twice of the longitudinal stress

That would be an indeterminate beam and the load distribution between the two, depends on the rigidity (EI) for each element. You may check this video for details: czcams.com/video/3kTMMNouHHo/video.html

@@MechanicsofMaterialsLibre Thank you. I have tried to complete my analysis but there is still some confusion. The analysis I am performing is as follows: Two circular hollow sections inside at different sizes that fixed to the ground and are overhang at the other end as shown in the video above. The question is: A 1.1m in height circular hollow section (CHS) is sized 42.40mm (diameter ) x 2.0mm (thickness ) with a 0.3m in height (CHS) inside that is sized 38.10mm (diameter) x 2.0mm (thickness ). There is a point load of 0.74kN acting on the top of the 1.1m tube, and it is assumed that both CHS are fixed at the bottom. How is the bending moment and deflection for the two posts calculated separately? I am finding it difficult to draw the deflection diagram since the smaller section would act to resist? How is the load distributed? by solving the slope equations?

Yes sigma_y is negative but here we used the absolute value since it is not changing the final answer.

The approximate area A_approx=πdt in thin walled pressure vessels (where d/t > 10) is close enough to the exact area A_exact = π/4*(d_outer^2-d_inner^2) and significantly simplifies the resultant stress equation. Note that 5% limit is usually considered sufficient in many engineering applications and such limit is satisfied in many thin walled pressure vessels. Thick walled pressure vessels have a non-uniform stress distribution and the calculation is totally different. Hope that helps.

@@MechanicsofMaterialsLibre Thank you for your explanation professor although I must strongly disagree with your approach. Where does it say exactly that there is an across the board 5% error limit that is acceptable in many engineering applications? Surely that would depend on a case by case basis and I doubt it would apply to error in your mathematical calculations. Is it then okay to say that 10 + 10 = 19 if, let us say, I am designing a nuclear reactor? How do you fit a 20 mm nut into a 19 mm bolt, it is only a 5% error? I get it, sometimes engineers have to accept a certain margin of error but surely not when you can help it?! Error margins are there for things like imperfections in materials, irregularities during manufacture and assembly, non-uniform stress distribution, factors that are impossible to accurately quantify. I do NOT believe error margins are there to excuse sloppy mathematical calculations purely for the sake of simplicity, especially not in an engineering dicipline. Frankly I would call that laziness and irresponsible. What happens when that unnecessary error is amplified when repeated in further calculations? What do you tell the Board of Enquiry when the structure collapses? "I made a few 5% errors because that was simplest but it was all within error margin."? As for simplifying the stress equation, yes I agree, multiplying 3 variables is MARGINALLY simpler than doing a simple area calculation for a circle, but the latter is not exactly rocket science either. This is basic elementary school mathematics we are talking about, you mean to tell me that doing an area calculation for a circle is too complex and that it is better to do an inaccurate calculation instead, purely for the sake of simplicity? I do not mean any disrespect but, either I do not understand the merit of your approach OR I simply cannot accept such a flagrant drop in standards. I think I will just keep doing the calculation the correct way for now. I recognise that you are an eminent professor in engineering and I am just a student so perhaps it is arrogant of me to say, but surely you can't be right?

This is a whole study in itself of tolerances. Anyhow you should know that while performing any calculations we are approximating. even in FEA, most fields have strict guidelines and rules like for steel structures in europe it EC3 and for piping its EN13480. Both are very conservative and in most cases will have a safety factor of 1.2 or more. I think you shouldn't be hard focused on how you got the #, every engineering problem you will work on will have some governing rules to define your #. Point of the tutorials is to get yourself familiar with procedures and approach to solutions.@@GryffieTube

ASME B&PV code uses precisely this approximation. THE PRESSURE FORCE EQUALS (w/4)D^2P WHILE THE RESISTING FORCE EQUALS (r)DTS. WHEN THE TWO ARE EQUATED, (v/4)D^2P = (w)DTS AND SOLVING FOR T = PD/AS = PR/2S. From European standard AD 2000 section B1 5 on the design of thin walled pressure vessels: The required wall thickness s is, in the case of spherical shells s = (D_a * p)/(40K/S * v + p) + c_1 + c_2 s wall thickness D_a out side diameter p design pressure K design strength S safety factor v Poisson's ratio Here expansion is approximated, but the same approximation is made. ASME B31.3 uses the same approximation when calculating stress in curved piping. This is not "@MechanicsofMaterialsLibre 's approach". This is an international design standard (based on barlow's formula) that has been used for decades in everything from rockets to nuclear power plants.

Barlow's formula is intrinsically an approximation(in far more ways than you have mentioned), and there are certainly more accurate ways of calculating stress in a pressure vessel. This video is about Barlow's method, if you wan't a more advanced technique, they really are not hard to find.